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Heat of 30 kcal is supplied to a system and 4200 J of external work is done on the sytem so that its volume decreases at constant pressure. What is the change in its internal energy $(J = 4200 J / k c a l)$

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$1.302 \times 10^{5} J$

$2.302 \times 10^{5} J$

$3.302 \times 10^{5} J$

$4.302 \times 10^{5} J$

answer is A.

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Detailed Solution

$Q = 30 k c a l, w = - 4200 J,$

$Δ Q = Δ U + Δ W \Rightarrow Δ U = Δ Q - Δ W$

$\Rightarrow ∆ U = 30 \times 4200 + 4200 = 1.302 \times 10^{5} J$

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